Question

at what height above the earth is the free-fall acceleration 40 % of its value at the surface? assume rearth = 6.37 Ý 106 m.

What is the speed of a satellite orbiting at that height? Assume Mearth = 5.98 Ý 1024 kg?

Answer 1

We know that the change in the gravity acceleration with height

g'=g(1-2h/R) ....1

here g'=40g/100 =0.4g

put these value in equation 1

0.4g=g(1-2h/R)

0.4=1-2h/R

2h/R=0.6

h=0.3R

Here R=6.37*10⁶ M

So

h=0.3*6.37*10⁶ =1.911*10⁶ m Answer.

We know that the centripetal force on the satellite is equal to the gravity force .

mv²/r =GMm/r²

here G=6.67*10^-11 , M=5.98*10²⁴ kg , r=R+h=R+0.3R=1.3R=8.281*10⁶

v=sqrt(6.67*10^-11*5.98*10²⁴/8.281*10⁶)

v=6.94*10³ m/s Answer